System of Equations and Inequalities

chapter 6 systems of equations and inequalities answers and system of equations and inequalities word problems
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11 Systems of Equations and Inequalities Figure 1 enigma machines like this one, once owned by Italian dictator benito mussolini, were used by government and military officials for enciphering and deciphering top-secret communications during World War II. (credit: dave Addey, Flickr) ChAPTeR OUTl Ine 11.1 Systems of l inear equations: Two variables 11.2 Systems of l inear equations: Three variables 11.3 Systems of nonlinear equations and Inequalities: Two variables 11.4 Partial Fractions 11.5 matrices and matrix Operations 11.6 Solving Systems with gaussian elimination 11.7 Solving Systems with Inverses 11.8 Solving Systems with Cramer's Rule Introduction By 1943, it was obvious to the Nazi regime that defeat was imminent unless it could build a weapon with unlimited destructive power, one that had never been seen before in the history of the world. In September, Adolf Hitler ordered German scientists to begin building an atomic bomb. Rumors and whispers began to spread from across the ocean. Refugees and diplomats told of the experiments happening in Norway. However, Franklin D. Roosevelt wasn’t sold, and even doubted British Prime Minister Winston Churchill’s warning. Roosevelt wanted undeniable proof. Fortunately, he soon received the proof he wanted when a group of mathematicians cracked the “Enigma” code, proving beyond a doubt that Hitler was building an atomic bomb. The next day, Roosevelt gave the order that the United States begin work on the same. e E Th nigma is perhaps the most famous cryptographic device ever known. It stands as an example of the pivotal role cryptography has played in society. Now, technology has moved cryptanalysis to the digital world. Many ciphers are designed using invertible matrices as the method of message transference, as finding the inverse of a matrix is generally part of the process of decoding. In addition to knowing the matrix and its inverse, the receiver must also know the key that, when used with the matrix inverse, will allow the message to be read. In this chapter, we will investigate matrices and their inverses, and various ways to use matrices to solve systems of equations. First, however, we will study systems of equations on their own: linear and nonlinear, and then partial fractions. We will not be breaking any secret codes here, but we will lay the foundation for future courses. 875876 CHAPTER 11 s t em os f e tq io N a s N d iN qe litise l eARnIng Obje CTIveS In this section, you will: • Solve systems of equations by graphing. • Solve systems of equations by substitution. • Solve systems of equations by addition. • Identify inconsistent systems of equations containing two variables. • Express the solution of a system of dependent equations containing two variables. 11.1 SySTemS OF lI ne AR eQUATIOnS: TWO vARIAble S Figure 1 (credit: Thomas Sørenes) A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a prot w fi ith its new line? How many skateboards must be produced and sold before a prot i fi s possible? In this section, we will consider linear equations with two variables to answer these and similar questions. Introduction to Systems of equations In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution. In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two die ff rent variables. For example, consider the following system of linear equations in two variables. 2x + y = 15 3x − y = 5 The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists. 2(4) + (7) = 15 True 3(4) − (7) = 5 True In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have ua ua ysSECTION 11.1 s sty ems of l i Near e aqu tio Ns: t o vw a riablse 877 different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satise fi s the system. u Th s, there are an infinite number of solutions. Another type of system of linear equations is an inconsistent system, which is one in which the equations represent two parallel lines. The lines have the same slope and die ff rent y-intercepts. There are no points common to both lines; hence, there is no solution to the system. types of linear systems e Th re are three types of systems of linear equations in two variables, and three types of solutions. • A n independent system has exactly one solution pair (x, y). The point where the two lines intersect is the only solution. • A n inconsistent system has no solution. Notice that the two lines are parallel and will never intersect. • A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations. Figure 2 compares graphical representations of each type of system. y y y 5 5 5 4 4 4 3 3 3 2 2 2 (1, 2) 7 11 , − 1 1 1 5 5 x 14 2 3 5 6 14 2 3 5 6 –6 –5 –4 –3 –2 –1 –6 –5 –4 –3 –2 –1 14 2 3 5 6 –6 –5 –4 –3 –2 –1 –1 –1 –1 –2 –2 (–1, –2) –2 –3 –3 –3 –4 –4 –4 –5 –5 –5 Independent System Inconsistent System Dependent System Figure 2 How To… Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution. 1. Substitute the ordered pair into each equation in the system. 2. Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution. Example 1 Determining Whether an Or dered Pair Is a Solution to a System of Equations Determine whether the ordered pair (5, 1) is a solution to the given system of equations. x + 3y = 8 2x − 9 = y Solution Substitute the ordered pair (5, 1) into both equations. (5) + 3(1) = 8 8 = 8 True 2(5) − 9 = (1) 1 = 1 True Th e ordered pair (5, 1) satise fi s both equations, so it is the solution to the system. Analysis We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines. See Figure 3.878 CHAPTER 11 s t em os f e tq io N a s N d iN qe litise y 5 4 3 2 (5, 1) 1 x +3y = 8 x –2 14 2 3 5 67 8 9 10 –1 –1 –2 2x 9 = y − –3 –4 –5 Figure 3 Try It 1 Determine whether the ordered pair (8, 5) is a solution to the following system. 5x − 4y = 20 2x + 1 = 3y Solving Systems of equations by graphing There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes. Example 2 Solving a System of Equations in T wo Variables by Graphing Solve the following system of equations by graphing. Identify the type of system. 2x + y = −8 x − y = −1 Solution Solve the first equation for y. 2x + y = −8 y = − 2x −8 Solve the second equation for y. x − y = −1 y = x + 1 Graph both equations on the same set of axes as in Figure 4. y 10 8 6 4 y = x + 1 y = −2x −8 2 x –5–4 –3 –2 –1 14 2 3 5 –2 –4 –6 (−3, −2) –8 –10 Figure 4 ua ua ysSECTION 11.1 s sty ems of l i Near e aqu tio Ns: t o vw a riablse 879 e l Th ines appear to intersect at the point ( −3, −2). We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations. 2(−3) + (−2) = −8 −8 = −8 True (−3) − (−2) = −1 −1 = −1 True e s Th olution to the system is the ordered pair ( −3, −2), so the system is independent. Try It 2 Solve the following system of equations by graphing. 2x − 5y = −25 −4x + 5y = 35 Q & A… Can graphing be used if the system is inconsistent or dependent? Yes, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system. Solving Systems of equations by Substitution Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical. How To… Given a system of two equations in two variables, solve using the substitution method. 1. Solve one of the two equations for one of the variables in terms of the other. 2. Substitute the expression for this variable into the second equation, then solve for the remaining variable. 3. Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair. 4. Check the solution in both equations. Example 3 Solving a System of Equations in Two Variables by Substitution Solve the following system of equations by substitution. −x + y = −5 2x − 5y = 1 Solution First, we will solve the first equation for y. −x + y = −5 y = x − 5880 CHAPTER 11 s t ems o f e tq io Ns a N d iN qe litise Now we can substitute the expression x − 5 for y in the second equation. 2x − 5y = 1 2x − 5(x − 5) = 1 2x − 5x + 25 = 1 −3x = −24 x = 8 Now, we substitute x = 8 into the first equation and solve for y. −(8) + y = −5 y = 3 Our solution is (8, 3). Check the solution by substituting (8, 3) into both equations. −x + y = − 5 −(8) + (3) = − 5 True 2x − 5y = 1 2(8) − 5(3) = 1 True Try It 3 Solve the following system of equations by substitution. x = y + 3 4 = 3x − 2y Q & A… Can the substitution method be used to solve any linear system in two variables? Yes, but the method works best if one of the equations contains a coec ffi ient of 1 or −1 so that we do not have to deal with fractions. Solving Systems of equations in Two variables by the Addition method A third method of solving systems of linear equations is the addition method. In this method, we add two terms with the same variable, but opposite coec ffi ients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coec ffi ients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition. How To… Given a system of equations, solve using the addition method. 1. Write both equations with x- and y-variables on the left side of the equal sign and constants on the right. 2. Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coec ffi ient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coec ffi ient of the same variable in the bottom equation, then add the equations to eliminate the variable. 3. Solve the resulting equation for the remaining variable. 4. Substitute that value into one of the original equations and solve for the second variable. 5. Check the solution by substituting the values into the other equation. ua ua ysSECTION 11.1 s sty ems of l i Near e aqu tio Ns: t o vw a riablse 881 Example 4 Solving a System by the Addition Method Solve the given system of equations by addition. x + 2y = −1 −x + y = 3 Solution Both equations are already set equal to a constant. Notice that the coec ffi ient of x in the second equation, −1, is the opposite of the coec ffi ient of x in the first equation, 1. We can add the two equations to eliminate x without needing to multiply by a constant. x + 2y = − 1 −x + y = 3 3y = 2 Now that we have eliminated x, we can solve the resulting equation for y. 3y = 2 2 _ y = 3 e Th n, w e substitute this value for y into one of the original equations and solve for x. −x + y = 3 2 __ −x + = 3 3 2 _ −x = 3 − 3 7 _ −x = 3 7 _ x = −  3 7 2 __ __ e s Th o lution to this system is −   ,   3 3 Check the solution in the first equation. x + 2y = −1 7 2 __ __ −   + 2 = −1     3 3 7 4 __ __ −   + = −1 3 3 3 __ −  = −1 3 −1 = −1 True Analysis We gain an important perspective on systems of equations by looking at the graphical representation. See Figure 5 to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph conr fi ms that the system has exactly one solution. y 5 4 7 2 x + 2y = −1 , 3 – 3 3 2 1 x –6 –5 –4 –3 –2 –1 14 2 3 56 –1 –2 −x + y = 3 –3 –4 –5 Figure 5882 CHAPTER 11 s t em os f e tq io Ns a N d iN qe litise Example 5 Using the Addition Method When Multiplication of One Equation Is Required Solve the given system of equations by the addition method. 3x + 5y = −11 x − 2y = 11 Solution Adding these equations as presented will not eliminate a variable. However, we see that the first equation has 3x in it and the second equation has x. So if we multiply the second equation by −3, the x-terms will add to zero. x − 2y = 11 −3(x − 2y) = −3(11) Multiply both sides by −3. −3x + 6y = −33 Use the distributive property. Now, let’s add them. 3x + 5y = −11 −3x + 6y = −33 11y = −44 y = −4 For the last step, we substitute y = −4 into one of the original equations and solve for x. 3x + 5y = − 11 3x + 5( − 4) = − 11 3x − 20 = − 11 3x = 9 x = 3 Our solution is the ordered pair (3, −4). See Figure 6. Check the solution in the original second equation. x − 2y = 11 (3) − 2( − 4) = 3 + 8 11 = 11 True y 6 5 4 3 2 1 x –6 –5 –4 –3 –2 –1 14 2 3 5 6 –1 –2 x − 2y = 11 –3 (3, –4) –4 –5 3x + 5y = −11 –6 Figure 6 Try It 4 Solve the system of equations by addition. 2x − 7y = 2 3x + y = −20 Example 6 Using the Addition Method When Multiplication of Both Equations Is Required Solve the given system of equations in two variables by addition. 2x + 3y = −16 5x − 10y = 30 ua ua ysSECTION 11.1 s sty ems of l i Near e aqu tio Ns: t o vw a riablse 883 Solution One equation has 2x and the other has 5x. The least common multiple is 10 x so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate x by multiplying the first equation by −5 and the second equation by 2. − 5(2x + 3y) = −5(−16) − 10x − 15y = 80 2(5x − 10y) = 2(30) 10x − 20y = 60 e Th n, we add the two equations together. −10x − 15y = 80 10x − 20y = 60 −35y = 140 y = −4 Substitute y = −4 into the original first equation. 2x + 3(−4) = −16 2x − 12 = −16 2x = −4 x = −2 Th e solution is (−2, −4). Check it in the other equation. 5x − 10y = 30 5(−2) − 10(−4) = 30 −10 + 40 = 30 30 = 30 See Figure 7. y 5 4 3 2 1 x –6 –5 –4 –3 –2 –1 14 2 3 5 6 –1 2x + 3y = −16 –2 –3 (−2, −4) –4 –5 5x − 10y = 30 –6 –7 Figure 7 Example 7 Using the Addition Method in Systems of Equations Containing Fractions Solve the given system of equations in two variables by addition. x y __ __ + = 3 3 6 x y __ __ − = 1 2 4 Solution First clear each equation of fractions by multiplying both sides of the equation by the least common denominator. x y __ __ 6 + = 6(3)   3 6 2x + y = 18 x y __ __ 4 − = 4(1)   2 4 2x − y = 4884 CHAPTER 11 s t em os f e tq io N a s N d iN qe litise Now multiply the second equation by −1 so that we can eliminate the x-variable. −1(2x − y) = −1(4) −2x + y = −4 Add the two equations to eliminate the x-variable and solve the resulting equation. 2x + y = 18 −2x + y = −4 2y = 14 y = 7 Substitute y = 7 into the first equation. 2x + (7) = 18 2x = 11 11 ___ x = 2 = 5.5 11 ___ e s Th olution is , 7 . Check it in the other equation.   2 x y __ __ − = 1 2 4 11 _ 7 2 ____ __ − = 1 2 4 11 7 ___ __ − = 1 4 4 4 __ = 1 4 Try It 5 Solve the system of equations by addition. 2x + 3y = 8 3x + 5y = 10 Identifying Inconsistent Systems of equations Containing Two variables Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but die ff rent y-intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as 12 = 0. Example 8 Solving an Inconsistent System of Equations Solve the following system of equations. x = 9 − 2y x + 2y = 13 Solution We can approach this problem in two ways. Because one equation is already solved for x, the most obvious step is to use substitution. x + 2y = 13 (9 − 2y) + 2y = 13 9 + 0y = 13 9 = 13 Clearly, this statement is a contradiction because 9 ≠ 13. Therefore, the system has no solution. ua ua ysSECTION 11.1 s sty ems of l i Near e aqu tio Ns: t o vw a riablse 885 The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows. x = 9 − 2y 2y = − x + 9 1 9 __ __ y = −   x + 2 2 We then convert the second equation expressed to slope-intercept form. x + 2y = 13 2y = − x + 13 1 13 __ ___ y = −   x + 2 2 Comparing the equations, we see that they have the same slope but different y-intercepts. Therefore, the lines are parallel and do not intersect. 1 9 __ __ y = −  x + 2 2 1 13 __ ___ y = −   x + 2 2 Analysis Writing the equations in slope-intercept form conr fi ms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown in Figure 8. y 12 10 8 1 9 y = − x + 2 2 6 4 1 13 y = − x + 2 2 2 x –12–10–8–6 –4 –2 28 4 6 10 12 –2 –4 –6 –8 –10 –12 Figure 8 Try It 6 Solve the following system of equations in two variables. 2y − 2x = 2 2y − 2x = 6 expressing the Solution of a System of dependent equations Containing Two variables Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as 0 = 0. Example 9 Finding a Solution to a Dependent System of Linear Equations Find a solution to the system of equations using the addition method. x + 3y = 2 3x + 9y = 6 Solution With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating x. If we multiply both sides of the first equation by −3, then we will be able to eliminate the x-variable.886 CHAPTER 11 s t em os f e tq io N a s Nd iN qe litise x + 3y = 2 (−3)(x + 3y) = (−3)(2) −3x − 9y = − 6 Now add the equations. −3x − 9y = −6 + 3x + 9y = 6 0 = 0 We can see that there will be an infinite number of solutions that satisfy both equations. Analysis If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let’s look at what happens when we convert the system to slope-intercept form. x + 3y = 2 3y = − x + 2 1 2 __ __ y = − x + 3 3 3x + 9y = 6 9y = −3x + 6 3 6 __ __ y = − x + 9 9 1 2 __ __ y = − x + 3 3 1 2 _ _ See Figure 9. Notice the results are the same. The general solution to the system is x, − x + .   3 3 y 5 4 3 2 x + 3y = 2 1 x –5 –4 –3 –2 –1 14 2 3 5 –1 3x + 9y = 6 –2 –3 –4 –5 Figure 9 Try It 7 Solve the following system of equations in two variables. y − 2x = 5 −3y + 6x = −15 Using Systems of equations to Investigate Profits Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation R = xp, where x = quantity and p = price. The revenue function is shown in orange in Figure 10. e Th cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in Figure 10. The x-axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars. ua ua ysSECTION 11.1 s sty ems of l i Near e aqu tio Ns: t o vw a riablse 887 70 60 Prot 50 40 (7, 33) 30 Cost Break-even 20 10 Revenue 0 –10 05 10 15 20 Quantity (in hundreds of units) Figure 10 e p Th oint at which the two lines intersect is called the break-even point. We can see from the graph that if 700 units are produced, the cost is 3,300 and the revenue is also 3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money. The shaded region to the right of the break-even point represents quantities for which the company makes a prot fi . The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as P(x) = R(x) − C(x). Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses. Example 10 Finding the Break-Even Point and the Profit Function Using Substitution Given the cost function C(x) = 0.85x + 35,000 and the revenue function R(x) = 1.55x, find the break-even point and the profit function. Solution Write the system of equations using y to replace function notation. y = 0.85x + 35,000 y = 1.55x Substitute the expression 0.85x + 35,000 from the first equation into the second equation and solve for x. 0.85x + 35,000 = 1.55x 35,000 = 0.7x 50,000 = x e Th n, we substitute x = 50,000 into either the cost function or the revenue function. 1.55(50,000) = 77,500 e b Th reak-even point is (50,000, 77,500). e p Th rot f fi unction is found using the formula P(x) = R(x) − C(x). P(x) = 1.55x − (0.85x + 35, 000) = 0.7x − 35, 000 e p Th rot f fi unction is P(x) = 0.7x − 35,000. Analysis The cost to produce 50,000 units is 77,500, and the revenue from the sales of 50,000 units is also 77,500. To make a profit, the business must produce and sell more than 50,000 units. See Figure 11. We see from the graph in Figure 12 that the profit function has a negative value until x = 50,000, when the graph crosses the x-axis. Then, the graph emerges into positive y-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss. Money (in hundreds of dollars)888 CHAPTER 11 s t em os f e tq io N a s Nd iN qe litise 100,000 Profit 60,000 Prot 50,000 80,000 40,000 Break-even point Prot 30,000 (50,000, 77,500) P(x) = 0.7x − 35,000 60,000 20,000 10,000 0 40,000 Break-even point Cost –10,000 (50,000, 0) C(x) = 0.85x + 35,000 –20,000 20,000 Revenue –30,000 R(x) = 1.55x –40,000 0 0 20,000 40,000 60,000 80,000100,000 Quantity Quantity Figure 11 Figure 12 Example 11 Writing and Solving a System of Equations in Two Variables e c Th ost of a ticket to the circus is 25.00 for children and 50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is 70,000. How many children and how many adults bought tickets? Solution Let c = the number of children and a = the number of adults in attendance. e t Th otal number of people is 2,000. We can use this to write an equation for the number of people at the circus that day. c + a = 2,000 e Th revenue from all children can be found by multiplying 25.00 by the number of children, 25c. e r Th evenue from all adults can be found by multiplying 50.00 by the number of adults, 50a. The total revenue is 70,000. We can use this to write an equation for the revenue. 25c + 50a = 70,000 We now have a system of linear equations in two variables. c + a = 2,000 25c + 50a = 70,000 In the first equation, the coec ffi ient of both variables is 1. We can quickly solve the first equation for either c or a. We will solve for a. c + a = 2,000 a = 2,000 − c Substitute the expression 2,000 − c in the second equation for a and solve for c. 25c + 50(2,000 − c) = 70,000 25c + 100,000 − 50c = 70,000 − 25c = −30,000 c = 1,200 Substitute c = 1,200 into the first equation to solve for a. 1,200 + a = 2,000 a = 800 We fi nd that 1,200 children and 800 adults bought tickets to the circus that day. Try It 8 Meal tickets at the circus cost 4.00 for children and 12.00 for adults. If 1,650 meal tickets were bought for a total of 14,200, how many children and how many adults bought meal tickets? Access these online resources for additional instruction and practice with systems of linear equations. • Solving Systems of equations Using Substitution (http://openstaxcollege.org/l/syssubst) • Solving Systems of equations Using elimination (http://openstaxcollege.org/l/syselim) • Applications of Systems of equations (http://openstaxcollege.org/l/sysapp) –20,000 –10,000 0 10,000 20,000 30,000 40,000 50,000 60,000 70,000 80,000 90,000 100,000 110,000 120,000 Dollars Dollars profit ua ua ysSECTION 11.1 s ectio N e xercises 889 11.1 SeCTIOn exeRCISeS veRbAl 1. Can a system of linear equations have exactly two 2. If you are performing a break-even analysis for solutions? Explain why or why not. a business and their cost and revenue equations are dependent, explain what this means for the company’s prot m fi argins. 3. If you are solving a break-even analysis and get 4. If you are solving a break-even analysis and there a negative break-even point, explain what this is no break-even point, explain what this means signifies for the company? for the company. How should they ensure there is a break-even point? 5. Given a system of equations, explain at least two different methods of solving that system. Algeb RAIC For the following exercises, determine whether the given ordered pair is a solution to the system of equations. 6. 5x − y = 4 7. −3x − 5y = 13 8. 3x + 7y = 1 x + 6y = 2 and (4, 0) − x + 4y = 10 and (−6, 1) 2x + 4y = 0 and (2, 3) 9. −2x + 5y = 7 10. x + 8y = 43 2x + 9y = 7 and (−1, 1) 3x − 2y = −1 and (3, 5) For the following exercises, solve each system by substitution. 11. x + 3y = 5 12. 3x − 2y = 18 13. 4x + 2y = −10 14. 2x + 4y = −3.8 2x + 3y = 4 5x + 10y = −10 3x + 9y = 0 9x − 5y = 1.3 15. −2x + 3y = 1.2 16. x − 0.2y = 1 17. 3 x + 5y = 9 18. −3x + y = 2 −3x − 6y = 1.8 −10x + 2y = 5 30x + 50y = −90 12x − 4y = −8 1 1 1 3 __ __ __ __ 19. x + y = 16 20. − x + y = 11 2 3 4 2 1 1 1 1 __ __ __ __ x + y = 9 − x + y = 3 6 4 8 3 For the following exercises, solve each system by addition. 21. −2x + 5y = −42 22. 6x − 5y = −34 23. 5x − y = −2.6 24. 7x − 2y = 3 7x + 2y = 30 2x + 6y = 4 −4x − 6y = 1.4 4x + 5y = 3.25 5 1 1 2 25. −x + 2y = −1 26. 7x + 6y = 2 1 __ __ __ __ __ x + y = 0 x + y = 27. 28. 6 4 3 9 9 5x − 10y = 6 −28x − 24y = −8 1 1 43 1 4 1 __ __ ___ __ __ __ x − y = − − x + y = − 8 2 120 2 5 3 29. −0.2x + 0.4y = 0.6 30. −0.1x + 0.2y = 0.6 x − 2y = −3 5x − 10y = 1 For the following exercises, solve each system by any method. 5 55 ___ ___ 31. 5x + 9y = 16 32. 6x − 8y = −0.6 33. 5x − 2y = 2.25 34. x − y = − 12 12 x + 2y = 4 3x + 2y = 0.9 7x − 4y = 3 5 55 __ ___ −6x + y = 2 2890 CHAPTER 11 s t em os f e tq io Ns a Nd iN qe litise 7 1 1 1 1 7 __ __ __ __ __ __ 35. 7x − 4y = 36. 3x + 6y = 11 37. x − y = 2 38. x + y = 3 6 6 2 3 3 2x + 4y = 9 1 21 3 3 1 1 __ ___ ___ __ __ __ 2x + 4y = − x + y = −3 x + y = − 3 6 12 2 4 8 39. 2.2x + 1.3y = −0.1 40. 0.1x + 0.2y = 2 4.2x + 4.2y = 2.1 0.35x − 0.3y = 0 gRAPhICAl For the following exercises, graph the system of equations and state whether the system is consistent, inconsistent, or dependent and whether the system has one solution, no solution, or infinite solutions. 41. 3x − y = 0.6 42. −x + 2y = 4 43. x + 2y = 7 x − 2y = 1.3 2x − 4y = 1 2x + 6y = 12 44. 3x − 5y = 7 45. 3x − 2y = 5 x − 2y = 3 −9x + 6y = −15 TeChn Ol Ogy For the following exercises, use the intersect function on a graphing device to solve each system. Round all answers to the nearest hundredth. 46. 0.1x + 0.2y = 0.3 47. −0.01x + 0.12y = 0.62 48. 0.5x + 0.3y = 4 −0.3x + 0.5y = 1 0.15x + 0.20y = 0.52 0.25x − 0.9y = 0.46 49. 0.15x + 0.27y = 0.39 50. −0.71x + 0.92y = 0.13 −0.34x + 0.56y = 1.8 0.83x + 0.05y = 2.1 exTen SIOnS For the following exercises, solve each system in terms of A, B, C, D, E, and F where A – F are nonzero numbers. Note that A ≠ B and AE ≠ BD. 51. x + y = A 52. x + Ay = 1 53. Ax + y = 0 54. Ax + By = C 55. Ax + By = C x − y = B x + By = 1 Bx + y = 1 x + y = 1 Dx + Ey = F ReAl-W ORld A PPl ICATIOnS For the following exercises, solve for the desired quantity. 56. A stuffed animal business has a total cost of 57. A fast-food restaurant has a cost of production production C = 12x + 30 and a revenue function C(x) = 11x + 120 and a revenue function R(x) = 5x. R = 20x. Find the break-even point. When does the company start to turn a prot? fi 58. A cell phone factory has a cost of production 59. A musician charges C(x) = 64x + 20,000, where x C(x) = 150x + 10,000 and a revenue function is the total number of attendees at the concert. The R(x) = 200x. What is the break-even point? venue charges 80 per ticket. Aer h ft ow many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point? 60. A guitar factory has a cost of production C(x) = 75x + 50,000. If the company needs to break even aer 150 uni ft ts sold, at what price should they sell each guitar? Round up to the nearest dollar, and write the revenue function. ua ua ysSECTION 11.1 s ectio N e xercises 891 For the following exercises, use a system of linear equations with two variables and two equations to solve. 61. Find two numbers whose sum is 28 and difference 62. A number is 9 more than another number. Twice is 13. the sum of the two numbers is 10. Find the two numbers. 63. e s Th tartup cost for a restaurant is 120,000, and 64. A moving company charges a flat rate of 150, each meal costs 10 for the restaurant to make. If and an additional 5 for each box. If a taxi each meal is then sold for 15, aer h ft ow many service would charge 20 for each box, how many meals does the restaurant break even? boxes would you need for it to be cheaper to use the moving company, and what would be the total cost? 65. A total of 1,595 first- and second-year college 66. 276 students enrolled in a freshman-level chemistry students gathered at a pep rally. The number of class. By the end of the semester, 5 times the number freshmen exceeded the number of sophomores by of students passed as failed. Find the number of 15. How many freshmen and sophomores were in students who passed, and the number of students attendance? who failed. 67. er Th e were 130 faculty at a conference. If there were 68. A jeep and BMW enter a highway running east- 18 more women than men attending, how many of west at the same exit heading in opposite directions. each gender attended the conference? e j Th eep entered the highway 30 minutes before the BMW did, and traveled 7 mph slower than the BMW. Aer 2 h ft ours from the time the BMW entered the highway, the cars were 306.5 miles apart. Find the speed of each car, assuming they were driven on cruise control. 69. If a scientist mixed 10% saline solution with 60% 70. An investor earned triple the prots o fi f what she saline solution to get 25 gallons of 40% saline earned last year. If she made 500,000.48 total solution, how many gallons of 10% and 60% for both years, how much did she earn in prots fi solutions were mixed? each year? 71. An investor who dabbles in real estate invested 1.1 72. If an investor invests a total of 25,000 into two million dollars into two land investments. On the bonds, one that pays 3% simple interest, and the first investment, Swan Peak, her return was a 110% 7 __ other that pays 2 % interest, and the investor 8 increase on the money she invested. On the second earns 737.50 annual interest, how much was investment, Riverside Community, she earned 50% invested in each account? over what she invested. If she earned 1 million in prots, h fi ow much did she invest in each of the land deals? 73. If an investor invests 23,000 into two bonds, one 74. CDs cost 5.96 more than DVDs at All Bets Are Off that pays 4% in simple interest, and the other paying Electronics. How much would 6 CDs and 2 DVDs 2% simple interest, and the investor earns 710.00 cost if 5 CDs and 2 DVDs cost 127.73? annual interest, how much was invested in each account? 75. A store clerk sold 60 pairs of sneakers. The high-tops 76. A concert manager counted 350 ticket receipts the sold for 98.99 and the low-tops sold for 129.99. day aer a co ft ncert. The price for a student ticket was If the receipts for the two types of sales totaled 12.50, and the price for an adult ticket was 16.00. 6,404.40, how many of each type of sneaker were e r Th egister confirms that 5,075 was taken in. How sold? many student tickets and adult tickets were sold? 77. Admission into an amusement park for 4 children and 2 adults is 116.90. For 6 children and 3 adults, the admission is 175.35. Assuming a different price for children and adults, what is the price of the child’s ticket and the price of the adult ticket?892 CHAPTER 11 s t em os f e tq io Ns a Nd iN qe litise l eARnIng Obje CTIveS In this section, you will: • Solve systems of three equations in three variables. • Identify inconsistent systems of equations containing three variables. • Express the solution of a system of dependent equations containing three variables. 11.2 SySTemS OF lI ne AR eQUATIOnS: ThRee vARIAble S Figure 1 (credit: “elembis,” Wikimedia Commons) John received an inheritance of 12,000 that he divided into three parts and invested in three ways: in a money-market fund paying 3% annual interest; in municipal bonds paying 4% annual interest; and in mutual funds paying 7% annual interest. John invested 4,000 more in municipal funds than in municipal bonds. He earned 670 in interest the first year. How much did John invest in each type of fund? Understanding the correct approach to setting up problems such as this one makes finding a solution a matter of following a pattern. We will solve this and similar problems involving three equations and three variables in this section. Doing so uses similar techniques as those used to solve systems of two equations in two variables. However, finding solutions to systems of three equations requires a bit more organization and a touch of visual gymnastics. Solving Systems of Three equations in Three variables In order to solve systems of equations in three variables, known as three-by-three systems, the primary tool we will be using is called Gaussian elimination, named ae ft r the prolic G fi erman mathematician Karl Friedrich Gauss. While there is no definitive order in which operations are to be performed, there are specic g fi uidelines as to what type of moves can be made. We may number the equations to keep track of the steps we apply. The goal is to eliminate one variable at a time to achieve upper triangular form, the ideal form for a three-by-three system because it allows for straightforward back-substitution to find a solution ( x, y, z), which we call an ordered triple. A system in upper triangular form looks like the following: Ax + By + Cz = D Ey + Fz = G Hz = K The third equation can be solved for z, and then we back-substitute to find y and x. To write the system in upper triangular form, we can perform the following operations: 1. Interchange the order of any two equations. 2. Multiply both sides of an equation by a nonzero constant. 3. Add a nonzero multiple of one equation to another equation. ua ua ysSECTION 11.2 s sty ems of l i Near e aqu tio Ns: t hre e variablse 893 The solution set to a three-by-three system is an ordered triple (x, y, z). Graphically, the ordered triple defines the point that is the intersection of three planes in space. You can visualize such an intersection by imagining any corner in a rectangular room. A corner is defined by three planes: two adjoining walls and the floor (or ceiling). Any point where two walls and the floor meet represents the intersection of three planes. number of possible solutions Figure 2 and Figure 3 illustrate possible solution scenarios for three-by-three systems. • Systems that have a single solution are those which, aer e ft limination, result in a solution set consisting of an ordered triple (x, y, z). Graphically, the ordered triple defines a point that is the intersection of three planes in space. • Systems that have an infinite number of solutions are those which, aer e ft limination, result in an expression that is always true, such as 0 = 0. Graphically, an infinite number of solutions represents a line or coincident plane that serves as the intersection of three planes in space. • Systems that have no solution are those that, aer e ft limination, result in a statement that is a contradiction, such as 3 = 0. Graphically, a system with no solution is represented by three planes with no point in common. (a) (b) Figure 2 ( a)Three planes intersect at a single point, representing a three-by-three system with a single solution. ( b) Three planes intersect in a line, representing a three-by-three system with infinite solutions. (a) (b) (c) Figure 3 All three figures represent three-by-three systems with no solution. ( a) The three planes intersect with each other, but not at a common point. (b) Two of the planes are parallel and intersect with the third plane, but not with each other. ( c) All three planes are parallel, so there is no point of intersection. Example 1 Determining Whether an Ordered Triple Is a Solution to a System Determine whether the ordered triple (3, −2, 1) is a solution to the system. x + y + z = 2 6x − 4y + 5z = 31 5x + 2y + 2z = 13 Solution We will check each equation by substituting in the values of the ordered triple for x, y, and z. x + y + z = 2 (3) + (−2) + (1) = 2 True 6x − 4y + 5z = 31 6(3) − 4(−2) + 5(1) = 31 18 + 8 + 5 = 31 True894 CHAPTER 11 s t ems o f e tq io N a s N d iN qe litise 5x + 2y + 2z = 13 5(3) + 2(−2) + 2(1) = 13 15 − 4 + 2 = 13 True e o Th rdered triple (3, −2, 1) is indeed a solution to the system. How To… Given a linear system of three equations, solve for three unknowns. 1. Pick any pair of equations and solve for one variable. 2. Pick another pair of equations and solve for the same variable. 3. You have created a system of two equations in two unknowns. Solve the resulting two-by-two system. 4. Back-substitute known variables into any one of the original equations and solve for the missing variable. Example 2 Solving a System of Three Equations in Three Variables by Elimination Find a solution to the following system: x − 2y + 3z = 9 (1) −x + 3y − z = −6 (2) 2x − 5y + 5z = 17 (3) Solution e Th re will always be several choices as to where to begin, but the most obvious first step here is to eliminate x by adding equations (1) and (2). x − 2y + 3z = 9 (1) −x + 3y − z = −6 (2) y + 2z = 3 (3) e s Th econd step is multiplying equation (1) by −2 and adding the result to equation (3). These two steps will eliminate the variable x. −2x + 4y − 6z = −18 (1) multiplied by − 2 2x − 5y + 5z = 17 (3) − y − z = −1 (5) In equations (4) and (5), we have created a new two-by-two system. We can solve for z by adding the two equations. y + 2z = 3 (4) −y − z = − 1 (5) z = 2 (6) Choosing one equation from each new system, we obtain the upper triangular form: x − 2y + 3z = 9 (1) y + 2z = 3 (4) z = 2 (6) Next, we back-substitute z = 2 into equation (4) and solve for y. y + 2(2) = 3 y + 4 = 3 y = −1 Finally, we can back-substitute z = 2 and y = −1 into equation (1). This will yield the solution for x. x − 2(−1) + 3(2) = 9 x + 2 + 6 = 9 x = 1 ua ua ys